Frequency synthesizers employ a feedback loop to lock the output, f.sub.o, to the input, f.sub.i. A phase detector responds to a difference in phase between the divided down input f.sub.in /N and the divided down feedback output f.sub.o /M by producing the appropriate charge up or charge down command to a charge pump. The charge pump pumps up or down the low frequency integrating capacitor increasing or decreasing, respectively, the voltage on it. The change in voltage drives an oscillator control amplifier which in turn drives an oscillator to increase or decrease the frequency f.sub.o to null the difference at the phase detector. In order to avoid positive feedback and to stabilize the system a high frequency capacitor, C.sub.H, is disposed in parallel with the integrating capacitor and a resistor, R, is placed in series with it. Typically C.sub.H is small (approximately 50.sub.pf) and can be mounted on board the frequency synthesizer chip as can R.sub.1 (approximately 1 K.OMEGA.) but C.sub.L is large (approximately 20,000 pf) and cannot easily be located on board and so is usually mounted as an external capacitor. One attempt, to mount it on the chip, distributes it widely across the chip under power supplies, buses and other elements. This assumes a low noise environment and ample buses and components to provide the necessary area. Another approach to this problem recognizes that at low frequencies the impedance of C.sub.H is not significant, at high frequencies the impedance of C.sub.L is not significant, and at mid-frequencies the impedance is essentially R. Therefore the C.sub.L and the R.sub.1 C.sub.H components can be considered as a series network with the same current I.sub.CP through all of them. Thus each of those networks can be treated separately so long as the voltage across them and their time constants remain the same. Therefore one could substantially reduce the current (to 1 .mu.a for example) through the larger external capacitor C.sub.L and so dramatically reduce its size, down to 50 pf, for example.
However, reducing the current to low levels runs the risk of producing deadbands due to the switches not switching properly in the charge pump. In addition the possible coupling induces charge injection from the semiconductor switches which is as large or even larger than the switched current and out of phase so that it can cause instability.